Let `I_(n) =int _(1) ^(e^(2))(ln x)^(n) dx (x ^(2)),` then the value of `3I_(n)+nI_(n-1)` equals to:

Options

A. 0
B. `2e ^(2)`
C. `e ^(2)`
D. `1`

Answer

To solve the integral \( I_n = \int_{1}^{e^2} (\ln x)^n \cdot x^2 \, dx \) and find the value of \( 3I_n + nI_{n-1} \), we can use integration by parts.

### Step-by-step Solution:

1. **Integration by Parts Setup**:
We will use integration by parts where we let:
- \( u = (\ln x)^n \) (first function)
- \( dv = x^2 \, dx \) (second function)

Then, we need to find \( du \) and \( v \):
- \( du = n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \)
- \( v = \frac{x^3}{3} \)

2. **Applying Integration by Parts**:
Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I_n = \left[ (\ln x)^n \cdot \frac{x^3}{3} \right]_{1}^{e^2} - \int_{1}^{e^2} \frac{x^3}{3} \cdot n (\ln x)^{n-1} \cdot \frac{1}{x} \, dx \]
Simplifying the integral:
\[ I_n = \left[ \frac{(e^2)^3}{3} \cdot (2)^n - \frac{1^3}{3} \cdot (0)^n \right] - \frac{n}{3} \int_{1}^{e^2} x^2 (\ln x)^{n-1} \, dx \]
The first term evaluates to:
\[ \frac{e^6}{3} \cdot 2^n \]
The second term is \( I_{n-1} \):
\[ I_n = \frac{e^6}{3} \cdot 2^n - \frac{n}{3} I_{n-1} \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ 3I_n + nI_{n-1} = 2^n e^6 \]

### Final Result:
Thus, the value of \( 3I_n + nI_{n-1} \) is:
\[ \boxed{2^n e^6} \]